Equation of a common tangent to conics frac{{{x^2}}}{{{a^2}}}

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10-2. Parabola, Ellipse, Hyperbola

hard

The equation of a common tangent to the conics $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$ is

$x + y = {a^2} - {b^2}$

$x + y = \sqrt {{a^2} - {b^2}} $

$x - y = \sqrt {{a^2} - {b^2}} $

$x + y = \sqrt {{b^2} - {a^2}} $

Solution

(b) The slopes of tangent on conic is

${m_1} = \frac{{b{x_1}}}{{a\sqrt {x_1^2 – {a^2}} }}$ and ${m_2} = \frac{{a{x_1}}}{{b\sqrt {x_1^2 + {b^2}} }}$

But tangent is common, so that ${m_1} = {m_2}$

==> $\frac{{b{x_1}}}{{a\sqrt {x_1^2 – {a^2}} }} = \frac{{a{x_1}}}{{b\sqrt {x_1^2 + {b^2}} }}$

By this equation we can find the value of ${x_1}$ and then ${y_1}$. Putting these values in $y – {y_1} = {m_1}(x – {x_1})$, we get the required equation.

Standard 11

Mathematics

The tangent to the hyperbola $xy = c^2$ at the point $P$ intersects the $x-$ axis at $T$ and the $y-$ axis at $T'$. The normal to the hyperbola at $P$ intersects the $ x-$ axis at $N$ and the $y-$ axis at $N'$. The areas of the triangles $PNT$ and $PN'T' $ are $ \Delta$ and $ \Delta ' $ respectively, then $\frac{1}{\Delta }\,\, + \,\,\frac{1}{{\Delta '}}\,$ is

normal

The value of $m$, for which the line $y = mx + \frac{{25\sqrt 3 }}{3}$, is a normal to the conic $\frac{{{x^2}}}{{16}} – \frac{{{y^2}}}{9} = 1$, is

medium

At the point of intersection of the rectangular hyperbola $ xy = c^2 $ and the parabola $y^2 = 4ax$ tangents to the rectangular hyperbola and the parabola make an angle $ \theta $ and $ \phi $ respectively with the axis of $X$, then

normal

If the two tangents drawn on hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ in such a way that the product of their gradients is ${c^2}$, then they intersects on the curve

hard

For $0<\theta<\pi / 2$, if the eccentricity of the hyperbola $\mathrm{x}^2-\mathrm{y}^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is :